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(x^2+3x+2)(x^2+7x+12)
=(x2+x+2x+2)(x2+3x+4x+12)
=[x.(x+1)+2.(x+1)][x.(x+3)+4.(x+3)]
=(x+1)(x+2)(x+3)(x+4)
![](https://rs.olm.vn/images/avt/0.png?1311)
Phân tích đa thức thành nhân tử(tách hạng tử)
1)x^2+2x-3=x^2-x+3x-3=x(x-1)+3(x-1)=(x-1)(x+3)
2)x^2-5x+6=x^2-2x-3x+6=x(x-2)-3(x-2)=(x-2)(x-3)
3)x^2+7x+12=(x+3)(x+4)
4)x^2-x-12=(x-4)(x+3)
5)3x^2+3x-36=3[(x-3)(x+4)]
6)5x^2-5x-10=5[(x-2)(x+1) ]
7)3x^2-7x-6=(x-3)(3x+2)
8)4x^2+4x-3=4x^2+6x-2x-3=(2x-1)(2x+3)
9)8x^2-2x-3=8x^2+4x-6x-3=(4x-3)(2x+1)
1: \(x^2+2x-3=\left(x+3\right)\left(x-1\right)\)
2: \(x^2-5x+6=\left(x-2\right)\left(x-3\right)\)
3: \(x^2+7x^2+12x=4x\left(2x+3\right)\)
4: \(x^2-x-12=\left(x-4\right)\left(x+3\right)\)
5: \(3x^2+3x-36=3\left(x^2+x-12\right)=3\left(x+4\right)\left(x-3\right)\)
6: \(5x^2-5x-10=5\left(x^2-x-2\right)=5\left(x-2\right)\left(x+1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
(x^2+3x+2)(x^2+7x+12)+1
=(x2+x+2x+2)(x2+3x+4x+12)+1
=[x.(x+1)+2.(x+1)][x.(x+3)+4.(x+3)]+1
=(x+1)(x+2)(x+3)(x+4)+1
=[(x+1)(x+4)][(x+2)(x+3)]+1
=(x2+5x+4)(x2+5x+6)+1
=(x2+5x+4)[(x2+5x+4)+2]+1
=(x2+5x+4)2+2(x2+5x+4)+1
=(x2+5x+4+1)2
=(x2+5x+5)2
![](https://rs.olm.vn/images/avt/0.png?1311)
(x^2+3x+2)(x^2+7x+12)-24
=(x2+x+2x+2)(x2+3x+4x+12)-24
=[x.(x+1)+2.(x+1)][x.(x+3)+4.(x+3)]-24
=(x+1)(x+2)(x+3)(x+4)-24
=(x+1)(x+4)(x+2)(x+3)-24
=(x2+5x+4)(x2+5x+6)-24
Đặt t=x2+5x+4 ta được:
t.(t+2)-24
=t2+2t-24
=t2-4t+6t-24
=t.(t-4)+6.(t-4)
=(t-4)(t+6)
thay t=x2+5x+4 ta được:
(x2+5x+4-4)(x2+5x+4+6)
=(x2+5x)(x2+5x+10)
=x.(x+5)(x2+5x+10)
Vậy (x^2+3x+2)(x^2+7x+12)-24=x.(x+5)(x2+5x+10)
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a) x2 - 7x + 5 = ( x2 - 2 . 7/2 . x + 49 / 4 ) + 5 - 49 / 4
= (x - 7/2)^2 - 29/4
= (x - 7/2)^2 - (√ 29 / 2 )^2
= ( x - ( 7 + √ 29 / 2 )). ( x + ( 7 - √ 29 / 2 ))
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(M=\left(x^2+3x+2\right)\left(x^2+7x+12\right)+1=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)
\(=\left[\left(x+1\right)\left(x+4\right)\right].\left[\left(x+2\right)\left(x+3\right)\right]+1=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1\)
Đặt \(t=x^2+5x+5\) \(\Rightarrow M=\left(t-1\right)\left(t+1\right)+1=t^2-1+1=t^2\)
Vậy \(M=\left(x^2+5x+5\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 3x2 - 7x + 4
= 3x2 - 3x - 4x + 4
= 3x( x - 1 ) - 4( x - 1 )
= ( x - 1 )( 3x - 4 )
b) x2 - 6xy + 9y2 = ( x - 3y )2
c) x2 - 8x - 9
= x2 - 9x + x - 9
= x( x - 9 ) + ( x - 9 )
= ( x - 9 )( x + 1 )
a) 3x2 - 7x + 4
= 3x2 - 4x - 3x + 4
= (3x2 - 4x) - (3x - 4)
= x.(3x - 4) - (3x - 4)
= (3x - 4).(x - 1)
b) x2 - 6xy + 9y2
= x2 - 2.x.3y + (3y)2
= (x - 3y)2
c) x2 - 8x - 9
= x2 - 9x + x - 9
= (x2 - 9x) + (x - 9)
= x.(x - 9) + (x - 9)
= (x - 9).(x + 1)
![](https://rs.olm.vn/images/avt/0.png?1311)
x^2+7x+12
=x^2+4x+3x+12
=x(x+4)+3(x+4)
=(x+3)(x+4)
3x^2-8x+5
=3x^2-3x-5x+5
=3x(x-1)-5(x-1)
(3x-5)(x-1)